NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
K 1 and K 2 are the respective equilibrium constant for the reactions (i) X e F 6 ( g ) + H 2 O ( g ) ⇌ X e O F 4 ( g ) + 2 H F ( g ) ; K 1 (ii) X e O 4 ( g ) + X e F 6 ( g ) ⇌ X e O F 4 ( g ) + X e O 3 F 2 ; K 2 The equilibrium constant for the reaction X e O 4 ( g ) + 2 H F ( g ) ⇌ X e O 3 F 2 ( g ) + H 2 O ( g ) will be
Options
- AK 1 K 2 2
- BK 1 - K 2
- CK 1 K 2
- DK 2 K 1
Correct answer
D. K 2 K 1
Step-by-step solution
(i) X e F 6 ( g ) + H 2 O ( g ) ⇌ X e O F 4 ( g ) + 2 H F ( g ) ; K 1 (ii) X e O 4 ( g ) + X e F 6 ( g ) ⇌ X e O F 4 ( g ) + X e O 3 F 2 ; K 2 Required equation X e O 4 ( g ) + 2 H F ( g ) ⇌ X e O 3 F 2 ( g ) + H 2 O ( g ) It is obtained by subtracting (i) from (ii), hence equilibrium constant of the required equation will be K 2 K 1 .